Just Another Inequality: Show that: (|a|+|b|)×(|b|+|c|)×(|c|+|a|)≥8×|a×b×c|

[Carrots]

Just Another Inequality: Show that: (|a|+|b|)×(|b|+|c|)×(|c|+|a|)≥8×|a×b×c|

Show that: (|a|+|b|)×(|b|+|c|)×(|c|+|a|)≥8×|a×b×c|

​

Now, what I tried to do was just multiply everything "manually", and needless to say I got stuck.
In the book the problem is meant to be solved using the Invariant Principle, which I didn't really get the hang of. What ways are there to solve this though?
Thanks in advance!

 

[Denis]

(|a|+|b|)×(|b|+|c|)×(|c|+|a|)≥8×|a×b×c|

Hmmmm....is that not same as:

(a + b)(a + c)(b + c) ≥ 8abc where a,b,c are integers > 0 ?

 

[Dr.Peterson]

Show that: (|a|+|b|)×(|b|+|c|)×(|c|+|a|)≥8×|a×b×c|

​

Now, what I tried to do was just multiply everything "manually", and needless to say I got stuck.
In the book the problem is meant to be solved using the Invariant Principle, which I didn't really get the hang of. What ways are there to solve this though?
Thanks in advance!

Please state the Invariant Principle, as you were taught it, so we can help you apply it. I can't tell what it is from what you have said, or from a search for the term, because there is not enough context.

Denis' idea is a good start, definitely. I can see several ways to reformulate the inequality, such as a product of pairwise arithmetic means on the left, and the product of the numbers themselves on the right.

It can probably be worked out similarly to a couple inequalities that have been asked about just today, but you may have another method that would be nicer.

 

[Deleted member 4993]

Show that: (|a|+|b|)×(|b|+|c|)×(|c|+|a|)≥8×|a×b×c|

​

Now, what I tried to do was just multiply everything "manually", and needless to say I got stuck.
In the book the problem is meant to be solved using the Invariant Principle, which I didn't really get the hang of. What ways are there to solve this though?
Thanks in advance!

If a,b & c>0, then:

(a+b)/2 ≥√(ab)

do the same for other pairs of numbers and multiply those together.